Math help thread

[mushy]

I dont get  it, why should this be complicated or confusing?

How about this:
10+5x2/3x8-3=?

If you mean  (10+5x2)/(3x8-3)
i.e.
10+5x2
3x8-3
then = 20/21

however
10+5x(2/3)x8-3= 7+(80/3) = 33 2/3

[Hawke.]

3 = square number

N3A + N3B + N

Find A and B with Simultaneous equation

[Ben.]

3 = square number

N3A + N3B + N

Find A and B with Simultaneous equation

With one equation  :trust:

[Pandalink]

oh, then 33 2/3

Yea this is correct.

Also Cofi the one I gave was supposed to be confusing because if you follow PEMDAS or BODMAS to the letter you will get the wrong answer. People don't know or forget that the DM and the AS are actually equivalent and that you read them left to right if written as such.

[Jubin]

3 = square number

N3A + N3B + N

Find A and B with Simultaneous equation

N(3A+1)=-3NB => 3A+1=-3B => 3(A+B)=-1
A+B=-1/3

So infinite amount of correct answers. Unless, you specify what you mean by 3= square number.

[[SE]Dr_Pepper27]

PEMDAS

( ) exponets multi/divi add/sub (left from right) pretty simple and my math teacher told me not  to use it, but I still do and get shit correct.

Easy way to remember: Please Excuse My Dear Aunt Sally or Please Excuse My Dope Ass Swag

[Hawke.]

So infinite amount of correct answers. Unless, you specify what you mean by 3= square number.

N ^3 or like 2^2 = 4

[CM Daniel]

N(3A+1)=-3NB => 3A+1=-3B => 3(A+B)=-1
A+B=-1/3

So infinite amount of correct answers. Unless, you specify what you mean by 3= square number.

What you're doing isn't possible, since he never said that N3A + N3B + N = 0. He never gave an equation.

He just gave:
N3A + N3B + N

This is not an equation. However, this would be an equation:
N3A + N3B + N = 0

Hawke's assignment makes no sense whatsoever.

[Aimzz]

I need help with Physcis, it is an assignment regarding Kinetic energy;
Someone knows how to solve this? I tried using the formula Ek=mv^2/2
Also the speed has to be converted to m/s? If so then I guess 55/3.6=15.28m/s
Do I need to count the potential energy aswell?

A roller coaster trolley and its passengers have a mass 840 Kg. The trolley comes over the top of the first hill with a speed of 55km/h. The hill is 85.0 m above the ground. The trolley goes down the first hill and up the rest of the second hill 64.0 m above ground. Ignore the effects of frictional forces. What is the kinetic energy of the trolley at the top of the second hill?

Help 

[CM Daniel]

I need help with Physcis, it is an assignment regarding Kinetic energy;
Someone knows how to solve this? I tried using the formula Ek=mv^2/2
Also the speed has to be converted to m/s? If so then I guess 55/3.6=15.28m/s
Do I need to count the potential energy aswell?

A roller coaster trolley and its passengers have a mass 840 Kg. The trolley comes over the top of the first hill with a speed of 55km/h. The hill is 85.0 m above the ground. The trolley goes down the first hill and up the rest of the second hill 64.0 m above ground. Ignore the effects of frictional forces. What is the kinetic energy of the trolley at the top of the second hill?

Help 

a = situation on the top of hill #1
b = situation on the top of hill #2
v_a = 55 / 3.6 = 15.28 ms^-1
g = 9.81 ms^-2
h_a = 85.0 m
h_b = 64.0 m

Ignore frictional forces, means no loss of energy. So:
E_total,a = E_total,b
E_kin,a + E_g,a = E_kin,b + E_g,b
E_kin,b = E_kin,a + E_g,a - E_g,b
E_kin,b = 0.5mv_a^2 + mgh_a - mgh_b
E_kin,b = 0.5 * 840 * 15.28^2 + 840 * 9.81 * 85.0 - 840 * 9.81 * 64.0 = 271109 = 2.7 * 10^5 J

So the kinetic energy on the second hill is 2.7 * 10^5 J(oule).

[Alarba]

a = situation on the top of hill #1
b = situation on the top of hill #2
v_a = 55 / 3.6 = 15.28 ms^-1
g = 9.81 ms^-2
h_a = 85.0 m
h_b = 64.0 m

Ignore frictional forces, means no loss of energy. So:
E_total,a = E_total,b
E_kin,a + E_g,a = E_kin,b + E_g,b
E_kin,b = E_kin,a + E_g,a - E_g,b
E_kin,b = 0.5mv_a^2 + mgh_a - mgh_b
E_kin,b = 0.5 * 840 * 15.28^2 + 840 * 9.81 * 85.0 - 840 * 9.81 * 64.0 = 271109 = 2.7 * 10^5 J

So the kinetic energy on the second hills is 2.7 * 10^5 J(oule).

This.

[Aimzz]

a = situation on the top of hill #1
b = situation on the top of hill #2
v_a = 55 / 3.6 = 15.28 ms^-1
g = 9.81 ms^-2
h_a = 85.0 m
h_b = 64.0 m

Ignore frictional forces, means no loss of energy. So:
E_total,a = E_total,b
E_kin,a + E_g,a = E_kin,b + E_g,b
E_kin,b = E_kin,a + E_g,a - E_g,b
E_kin,b = 0.5mv_a^2 + mgh_a - mgh_b
E_kin,b = 0.5 * 840 * 15.28^2 + 840 * 9.81 * 85.0 - 840 * 9.81 * 64.0 = 271109 = 2.7 * 10^5 J

So the kinetic energy on the second hills is 2.7 * 10^5 J(oule).

Thanks for the help! 

[Jubin]

N ^3 or like 2^2 = 4

Ah, you know we have commands for typing it like that anyway then you'll get

3N²(A+B)= -1

[mushy]

Ah, you know we have commands for typing it like that anyway then you'll get

3N²(A+B)= -1

Christ thats a tuffy. Specially at this time of night.
I'll get it.

[CM Daniel]

Thanks for the help! 

No problem.

Now here's one for you guys, for fun.

(x^2 + 6)^9. Write out it (leaving no more brackets), and yes, you don't need a computer program to do so.